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Chapter 9: Problem 13

Use the window \([-2,2] \times[-2,2]\) to sketch a direction field for the following equations. Then sketch the solution curve that corresponds to the given initial condition. A detailed direction field is not needed. $$y^{\prime}(t)=4-y, y(0)=-1$$

Short Answer

Expert verified
Question: Sketch the direction field for the given differential equation $$y^{\prime}(t) = 4 - y$$ within the window \([-2, 2] 脳 [-2, 2]\), and find the solution for the initial condition y(0) = -1. Answer: The direction field can be drawn by plotting several short line segments with the slopes given by the differential equation $$y^{\prime}(t) = 4 - y$$ at various points within the given window. For the initial condition y(0) = -1, the solution is $$y(t) = 4 - 5e^t$$.

Step by step solution

01

Determine the window for the plot

The given window for the plot is \([-2, 2] 脳 [-2, 2]\), which means that we are going to create a direction field for the differential equation within this window.
02

Calculate the slope for the direction field

In order to sketch the direction field, we need to calculate the slopes at points within this window. For the given differential equation, $$y^{\prime}(t) = 4 - y$$, the slopes depend on the value of y.
03

Sketch the solution curve that matches the given initial condition

We are given that y(0) = -1 as an initial condition. To find the solution curve for this initial condition, we will solve the differential equation. $$ y^{\prime}(t) = 4 - y $$ $$ \frac{dy}{dt} = 4 - y $$ Isolating the y terms and integrating both sides: $$ \int \frac{dy}{4-y} = \int dt $$ For this, we make a substitution: \(u=4-y\Rightarrow du=-dy\) $$ -\int \frac{1}{u}du = \int dt $$ Now, we integrate both sides: $$ -\ln(u) = t + C_{1} $$ Replacing \(u\) with \((4-y)\): $$ -\ln(4-y) = t + C_{1} $$ To get the value of \(C_{1}\) and the explicit solution for y(t), we use the initial condition y(0)=-1: $$ -\ln(4 - (-1)) = 0 + C_{1} $$ $$ -\ln(5) = C_{1} $$ Now we have the constant, and we can write the solution: $$ -\ln(4-y) = t - \ln(5) $$ To make this easier to plot, we can exponentiate both sides and isolate y: $$ 4-y = e^{t-\ln(5)} $$ $$ y(t) = 4 - 5e^t $$
04

Draw the direction field

Now that we have the solution curve for the given initial condition, we can sketch the direction field for the window \([-2, 2] 脳 [-2, 2]\). To do this, we will simply plot several short line segments with the slopes given by the differential equation $$y^{\prime}(t) = 4 - y$$ at various points within the window. Additionally, we will sketch the solution curve that we found, $$ y(t) = 4 - 5e^t $$, onto the direction field to visualize the solution for the initial condition y(0) = -1. The resulting sketch will show the direction field and the solution curve for the given differential equation within the window \([-2, 2] 脳 [-2, 2]\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Differential Equations
Differential equations are a type of equation that involve an unknown function and its derivatives. They are a powerful tool in mathematics and science for modeling how quantities change over time or in response to various factors. The general form of a first-order differential equation is expressed as \( y^{\text{'} }(t) = f(t, y) \), where \( y \) is the unknown function of \( t \), and \( f(t, y) \) is a given function that relates \( y \) to its derivative.

When solving a differential equation, the goal is to find a function \( y(t) \) such that when the function and its derivative are substituted into the equation, the equation is satisfied. This exercise involves the first-order linear differential equation \( y^{\text{'} }(t)=4-y \), for which the solution will provide the particular way the variable \( y \) evolves over time.
Initial Value Problem
An initial value problem is a type of differential equation that comes with an additional piece of information: the value of the unknown function at a specific point, termed as the initial condition. This information is crucial because differential equations can have infinitely many solutions without it.

In our example, the initial condition is given by \( y(0)=-1 \). This tells us the value of the function \( y(t) \) when \( t = 0 \), which we use to solve for the constant of integration that arises when the equation is integrated. The inclusion of the initial condition allows us to find the specific solution to the differential equation that not only meets the equation's requirements but also passes through the point \( (0, -1) \) on a graph of \( y \) versus \( t \).
Slope Field
A slope field, also known as a direction field, is a visual representation of a first-order differential equation. It consists of a grid of points at which small line segments are drawn. The slope of each line segment corresponds to the value of the derivative \( y^{\text{'} }(t) \) at that point. Essentially, a slope field shows the direction in which the solution curve is headed at any point in the plane.

In practice, to sketch a slope field, one would compute the derivative value for several points within the given window, like \([-2, 2] \times [-2, 2]\), and then draw line segments with these slopes. Once the slope field is constructed, one can visualize how solution curves would traverse this field. Plotting the particular solution curve that corresponds to the given initial condition adds context to the general behavior suggested by the field.
Integrating Factor Method
The integrating factor method is a technique used to solve linear first-order differential equations. The aim of the method is to multiply the differential equation by a function, the integrating factor, which turns the left-hand side of the equation into the derivative of a product. This facilitates the integration of both sides of the equation.

For our equation \( y^{\text{'} }(t) = 4 - y \), though not directly applied in the solution steps provided, the integrating factor is \( e^{\text{-}t} \). If we applied this method, we would multiply both sides of the equation by this factor, which would result in an equation that could be integrated more straightforwardly. It's a powerful tool for solving a range of more complicated differential equations and understanding this method can simplify many problems in differential calculus.

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Most popular questions from this chapter

Convergence of Euler's method Suppose Euler's method is applied to the initial value problem \(y^{\prime}(t)=a y, y(0)=1,\) which has the exact solution \(y(t)=e^{a t} .\) For this exercise, let \(h\) denote the time step (rather than \(\Delta t\) ). The grid points are then given by \(t_{k}=k h .\) We let \(u_{k}\) be the Euler approximation to the exact solution \(y\left(t_{k}\right),\) for \(k=0,1,2, \ldots\). a. Show that Euler's method applied to this problem can be written \(u_{0}=1, u_{k+1}=(1+a h) u_{k},\) for \(k=0,1,2, \ldots\). b. Show by substitution that \(u_{k}=(1+a h)^{k}\) is a solution of the equations in part (a), for \(k=0,1,2, \ldots\). c. Recall from Section 4.7 that \(\lim _{h \rightarrow 0}(1+a h)^{1 / h}=e^{a} .\) Use this fact to show that as the time step goes to zero \((h \rightarrow 0,\) with \(\left.t_{k}=k h \text { fixed }\right),\) the approximations given by Euler's method approach the exact solution of the initial value problem; that is, \(\lim _{h \rightarrow 0} u_{k}=\lim _{h \rightarrow 0}(1+a h)^{k}=y\left(t_{k}\right)=e^{a t_{k}}\).

Explain how a stirred tank reaction works.

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Solve the differential equation for Newton's Law of Cooling to find the temperature function in the following cases. Then answer any additional questions. An iron rod is removed from a blacksmith's forge at a temperature of \(900^{\circ} \mathrm{C}\). Assume \(k=0.02\) and the rod cools in a room with a temperature of \(30^{\circ} \mathrm{C}\). When does the temperature of the rod reach \(100^{\circ} \mathrm{C} ?\)

The following initial value problems model the payoff of a loan. In each case, solve the initial value problem, for \(t \geq 0,\) graph the solution, and determine the first month in which the loan balance is zero. $$B^{\prime}(t)=0.004 B-800, B(0)=40,000$$
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